I have a question on Feynman-Kac formula but can I ask the community if I have done it correctly? If no, may you point out to where I went wrong? Thanks!

The original FK formula states: Assume $f(t,x)$ satisfies $$\frac{\partial f}{\partial t}(t,x) + \mu(t,x)\frac{\partial f}{\partial x} + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t,x)\sigma^2(t,x) = 0$$, $f(T,x) = \phi(x)$. Then $f(s,x) = E_{s,x} [\phi(X_T)]$ where $dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t$.

Btw, $E_{s,x}[.]$ denotes the conditional expectation $E[.|\mathcal F_s]$ and $X_s = x$.

The given question I have is: Find solution $F(t,x)$ to $$\frac{\partial f}{\partial t} + \mu x\frac{\partial f}{\partial x} + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}\sigma^2x^2 = 0$$ with terminal condition $F(T,x) = \ln(x^3)$

My solution is to choose an appropriate $X_t$ that satisfies the dynamics of $X_t$ in the FK formula. In this case, I believe the choice is the Geometric Brownian Motion that has solution $X_t = X_s e^{(\mu - \frac{\sigma^2}{2})(t-s) + \sigma (W_t-W_s)}$ then \begin{align}F(t,x) &= E_{t,x} [\ln(X_T^3)] \\ &= E_{t,x} \left[ 3\ln(x) + 3(\mu - \frac{\sigma^2}{2})(T-t) + 3\sigma (W_T-W_t)\right] \end{align} So far, am I right? But the last term equals $0$ since $E[W_T - W_t | \mathcal F_t] = E[W_T - W_t]$.

  • 2
    $\begingroup$ Because Brownian motion is a martingale, $E[W_T-W_t|\mathcal{F}_t]=W_t-W_t=0$ $\endgroup$
    – Kevin
    Sep 21, 2020 at 14:36
  • $\begingroup$ yes, I overlooked that. $\endgroup$ Sep 21, 2020 at 17:11


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